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If a, b, c are sides of triangle. such that $3 (a^{2} + b^{2} + c^{2} + 1) - 2 (a + b + c) = 2 (a b + b c + c a)$ then the area of the triangle is

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$\frac{3 \sqrt{3}}{4}$

$\frac{4 \sqrt{3}}{5}$

$\frac{\sqrt{3}}{4}$

$\sqrt{3}$

answer is C.

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Detailed Solution

$2 (a^{2} + b^{2} + c^{2} - a b - b c - c a) + (a^{2} + b^{2} + c^{2} - 2 a - 2 b - 2 c + 3) = 0$

${(a + b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} + {(a - 1)}^{2} + {(b - 1)}^{2} + {(c - 1)}^{2} = 0$

$∴ a = b = c = 1$

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